POJ 3233 Matrix Power Series

本题地址：>here<

Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

Source

POJ Monthly–2007.06.03, Huang, Jinsong

题解

很明显这是个水题
就是那个…矩阵的累乘和
自行百度（原谅我不会搞数学符号）
时间复杂度是O（n³log k）

#include <cstdio>
#include <vector>
using namespace std;
typedef vector<int> vec;
typedef vector<vec> mat;

int m;
mat mul(mat &A,mat &B) {
    mat C(A.size(),vec(B[0].size()));
    for(int i=0;i<A.size();i++)
        for(int k=0;k<B.size();k++)
            for(int j=0;j<B[0].size();j++)
                C[i][j]=(C[i][j]+A[i][k]*B[k][j])%m;
    return C;
}

mat pow(mat A,int n) {
    mat B(A.size(),vec(A.size()));
    for(int i=0;i<A.size();i++)
        B[i][i]=1;
    while(n>0) {
        if(n&1) B=mul(B,A);
        A=mul(A,A);
        n>>=1;
    }
    return B;
} 

int n,k;
mat A;
int main() {
    scanf("%d%d%d",&n,&k,&m);
    mat B(n*2,vec(n*2));
    for(int i=0;i<n;i++) {
        vector<int> a;
        for(int j=0;j<n;j++) {
            int t;
            scanf("%d",&t);
            a.push_back(t);
        }
        A.push_back(a);
    }
    for(int i=0;i<n;i++) {
        for(int j=0;j<n;j++)
            B[i][j]=A[i][j];
        B[n+i][i]=B[n+i][n+i]=1;
    }
    B=pow(B,k+1);
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++) {
            int a=B[n+i][j]%m;
            if(i==j) a=(a+m-1)%m;
            printf("%d%c",a,j==n-1?'\n':' ');
        }
    return 0;
}